\hspace{-16}$I Dont have much Idea about that type of Integral....\\\\\\ May be it is an Elleptical Integral of $\bf{(I)}$ or $\bf{(II)}$ Type..\\\\\\Given $\bf{\int\frac{x^2}{\sqrt{1-x^4}}\;dx = \int x^2\cdot \left(1-x^4\right)^{-\frac{1}{2}}dx}$\\\\\\ Now Using Chebyshev integral....\\\\\\ $\bf{\bullet \;\; \int x^m\cdot (a+bx^n)^{p}\; dx}$\\\\\\ Where $\bf{m,n,p}$ are Rational no. and the given Integral\\\\\\ is Integral in elementry form only if ............\\\\\\ $\bf{\left(\frac{m+1}{n}\right)\in \mathbb{Z}}$ \bf{\underline{\underline{or}}} $\bf{\left(\frac{m+1}{n}+p\right)\in \mathbb{Z}}$\\\\\\ Now in $\bf{\int x^2\cdot \left(1-x^4\right)^{-\frac{1}{2}}dx}$\\\\\\ Here $\bf{m=2\;\;,a=1\;\;,b=-1\;\;,n=4\;\;,p=-\frac{1}{2}}$\\\\\\
\hspace{-20}$So $\bf{\left(\frac{2+1}{4}\right)\neq \mathbb{Z}}$ \bf{\underline{\underline{or}}} $\bf{\left\{\left(\frac{2+1}{4}-\frac{1}{2}\right)\right\}\neq \mathbb{Z}}$\\\\\\ So we can Not Integrate $\bf{\int\frac{x^2}{\sqrt{1-x^4}}\;dx}$ in terms of\\\\\\ elementry function..
Aditya Agrawal Can you provide a place where I can find these kind of questions? Maybe Chebyshev Integral Questions?
Upvote·0· Reply ·2014-06-24 20:11:04Aditya Agrawal And how do you write your answers in this kind of form?
man111 singh Using Online Latex Editor... Many Question Give in Preparation Books and Coaching Meterials.. Like Integration of 1/(1+x^4)^(1/4) or 1/(1-x^4)^(1/4) or (x)^(-1/2){(3+x^(1/3))^2} or (x)^(-1/2){(2+3x^(1/3))^(-2)} or x^(1/3).{1+x^(4/3)}^(1/7) or (x)^(-6){(1+2x^(3))^(2/3)}